$\begingroup$OP does not show any attempt not to advance research and previous work. For questions so people give negative feedback but why now do not? There are also cases where the question is put on hold and now they do not. Can, please be more consistent?$\endgroup$
– Luis FelipeOct 22 '15 at 15:46

31 Answers
31

One way to find the coefficients, assuming we already know that it's a degree $3$ polynomial, is to calculate the sum for $n=0,1,2,3$. This gives us four values of a degree $3$ polynomial, and so we can find it.

The better way to approach it, though, is through the identity
$$ \sum_{t=0}^n \binom{t}{k} = \binom{n+1}{k+1}. $$
This identity is true since in order to choose a $(k+1)$-subset of $n+1$, you first choose an element $t+1$, and then a $k$-subset of $t$.

$\begingroup$That's a good method: +1! Concerning the polynomial approach, I think it could be enhanced by noticing that if $p_k(n)=\sum_{i=1}^n i^k$ then, extending what you wrote, $p_k\in\mathbb Q[n]$, $\partial p_k=k+1$, and $n(n+1)\,\Big|\,p_k$ for every $k$. For $k=2$ this reduces to 2 the coefficients to be found making this approach equally worth, isn't it?$\endgroup$
– AndreasTMar 30 '13 at 15:58

$\begingroup$That's a great approach. Can you tell me what's the name of this identity : $\sum_{t=0}^n \binom{t}{k} = \binom{n+1}{k+1}$. Is it the Pascal triangle?$\endgroup$
– hlapointeOct 21 '15 at 14:37

1

$\begingroup$@hlapointe It probably does have a name but I can't think of any! Do let me know if you find out.$\endgroup$
– Yuval FilmusOct 21 '15 at 15:27

I think it's useful to report here another proof that I have posted on Mathoverflow.

Write down numbers in an equilateral triangle as follows:

1
2 2
3 3 3
4 4 4 4

Now, clearly the sum of the numbers in the triangle is $Q_n:=1^2+2^2+\dots+n^2$. On the other hand, if you superimpose three such triangles rotated by $120^\circ$ each, like these ones

1 4 4
2 2 3 4 4 3
3 3 3 2 3 4 4 3 2
4 4 4 4 1 2 3 4 4 3 2 1

then the sum of the numbers in each position equals $2n+1$. Therefore, you can double-count $3Q_n=\frac{n(n+1)}{2}(2n+1)$. $\square$

The proof is not mine and I do not claim otherwise. I first heard it from János Pataki. It is similar (but simpler) to the proof appearing on Wikipedia as I am writing this.

How to prove formally that all positions sum to $2n+1$? Easy induction: moving down-left or down-right from the topmost number does not alter the sum, since one of the three summand increases and one decreases. This is a discrete analogue of the Euclidean geometry theorem "given a point $P$ in an equilateral triangle $ABC$, the sum of its three distances from the sides is constant" (proof: sum the areas of $APB,BPC,CPA$), which you can mention as well.

To discover the identity, notice that any polynomial solution of the above recurrence has degree at most $3$. Hence it's easy to find the polynomial solution by substituting a cubic polynomial with undetermined coefficients.

Similarly, we can derive for each $k$
$$ \sum_{j=0}^{k-1} {k \choose j} E(X^j) = \sum_{l=1}^k {k \choose l} n^{l-1} $$
and so if we know $E(X^0), \ldots, E(X^{k-2})$ we can solve for $E(X^{k-1})$. So this method generalizes to higher moments as well.

Sums of polynomials can be done completely mechanically (no insight required, just turn the handle!) using the discrete calculus. Bill Dubuque mentions this in his answer, but I think it's nice to see a worked example.

A natural approach for this kind of problems when you don't know the result is to proceed as follows :

We may want to write the sum $\sum_{k=1}^n k^2$ as a telescopic sum, so we will try to find a polynomial of degree 3 ( why ? ) $P$ so that $P\left( k+1 \right) - P\left(k\right)=k^2$. Let $P\left( x \right) = ax^3+bx^2+cx$ for all reals $x$, then our constraint becomes :

Let $S=\{1,2,\dots,(n+1)\},n\ge 2$ and $T=\{(x,y,z)|x.y,z\in S,x< z,y< z\}$.By counting the number of members of $T$ in $2$ different ways I will prove the formula.

$1$st way:

We will at first Choose $z$ form the set $S$.When $z$ is $1$ then there are no choices for $x,y$ so the no. of elements of $T$ with $z=0$ is zero.When $z=2$ the number of choices for $x$ is $1$ and so is for $y$(precisely $x=y=1$).When $z=3$ then $x\in \{1,2\}$ and $y\in \{1,2\}$ so total no. of choices equals $2^2$.In a similar manner when $z=k,(1\le k\le (n+1))$,no. of choices for $x$ equals $(k-1)$ and no. of choices for $y$ is also $(k-1)$. So total no . of elements of T with $z=k$ is $(k-1)^2$.

So we will get the total no. of elements of $T$ by summing $(k-1)^2$ up from $1 $ to $(n+1)$.Hence $$|T|=\sum _{l=1}^{(n+1)}(l-1)^2=\sum_{k=1}^{n}k^2$$

$2$nd way:

Among the elements of $T$ consisting of three numbers from the set $S$, there are elements in $x=y$ and elements in which $x\ne y$.

We can count the no. of elements in which $x=y$ by choosing two distinct nos. from $S$ and assigning $z$ with the lagest no. and $x,y$ with the smallest number. We can choose two distinct numbers from $S$ in $\displaystyle \binom{n+1}{2}$ ways, so the total no. elements having $x=y$ is $\displaystyle \binom{n+1}{2}$.

Now we have to count the number of elements in which $x\ne y$.This means that $x,y$ are dinstict and as they are less than $z$ this means that all the three are distinct. So we can count no. of such elements in $T$ in the following way.At first we will choose three elements from the set $S$ and assign the largest value to $z$ and assign the other two values to $x,y$. Now we can choose three numbers from the set $S$ in $\displaystyle \binom{n+1}{3}$.From each such three element we can get two elements of the set $T$(Assigning the largest to $z$ and then assigning any one of then to $x$ and the other to $y$). So no. of elements of $T$ having $x\ne y$ is $2\displaystyle \binom{n+1}{3}$

So by this method we have $|T|=\displaystyle \binom{n+1}{2}+2\displaystyle \binom{n+1}{3}$

On equating the result obtained from both the methods we have
$$\sum_{k=1}^{n}k^2=\displaystyle \binom{n+1}{2}+2\displaystyle \binom{n+1}{3}=\frac{n(n+1)(2n+1)}{6}$$

Note that this can easily be extended to find the sum of $p$th power of integers.($p\in \mathbb{N})$

$\begingroup$@user236182 - The following is a well-known identity: $$\sum_{k=m}^N\binom km=\binom {N+1}{m+1}$$ A proof of it may be found here as a solution to a recent question. Putting $N=2n+1$ results in the identity used in the solution.$\endgroup$
– hypergeometricOct 22 '15 at 14:42

However, if you're like most people, inverting a $4\times4$ matrix doesn't exactly tickle your fancy!

Luckily, now that we see that the interpolating cubic is unique, we could find it through the described matrix multiplication, but we would get to the same result if we proceeded a different route as well. This is where Lagrange polynomials come to the rescue.

Base case: If $n=0$, then we have $0$ on the left hand side, and $0(0+1)(2(0)+1)/6=0$ on the right.

Induction step:

Consider the differences $L(j+1)-L(j)$, and $R(j+1)-R(j)$ where $L(j)$ indicates that we have $j$ for $n$ on the left hand side. Well, $L(j+1)-L(j)=(j+1)^2$, and
$$R(j+1)-R(j)=\frac{(j+1)((j+1)+1))(2(j+1)+1)}{6} - \frac{j(j+1)(2j+1)}{6}$$ which simplifies to $(j+1)^2$ also. So, the rates of change on both sides equal each other, and thus the induction step follows.

Which is the relation we set out to prove. So the method is to substitute $i=k+1$ into the formula you are trying to prove and then use the inductive assumption to recover the $\color{blue}{\mathrm{blue}}$ equation at the end.

$f(x) = x^2$ is a strictly monotonously increasing function. The sum is an upper Riemann sum of width 1 of this function. But if we shift it down one step we get a lower Riemann sum of width 1. The difference in integral is $$\int_{n-1}^n f(x)dx - \int_0^1 f(x)dx = n^3/3 - (n-3)^3/3 - 1 = n^2-3n-4$$

This is not quite $n^2$, but we can now turn the question into asking "if we had an offset in the integration, say $+\alpha$ on each "step", what $\alpha$ should we select to get as close to $n^2$ as possible?"

This is not a full answer, but intends on helping how to think to try and solve problems.

If you know that the nth square is the sum of the first n odd numbers, you can rewrite each square in the above sum in that way and do a little bit of rearranging to get the desired identity. In general, knowing the value of (n + 1)^k - n^k allows you to write (n+1)^k as a telescoping sum of a polynomial of degree k-1, running over the values 1 through n. If you know the values of the sums of consecutive powers up to k-1, this allows you to find the sum of consecutive kth powers by substituting the polynomial sum for each kth power.

Thank you for your interest in this question.
Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).